1. Field of the Invention
The invention generally relates to wired digital communications systems. In particular, the invention relates to the equalization of signals that can be severely distorted by non-linearities.
2. Description of the Related Art
Any of a variety of physical impairments can limit the effective transmission of data signals over communications channels. For example, the frequency selective nature of the channels can cause different frequency components of the input signal to be attenuated and phase-shifted differently. This can cause the impulse response of the channel to span several symbol intervals, resulting in time-smearing and interference between successive transmitted input symbols, commonly known as intersymbol interference (ISI). The ISI resulting from the channel distortion, if left uncompensated, can cause high error rates. The solution to the ISI problem is to design a receiver that compensates for the ISI in the received signal. The compensator for ISI is known as an equalizer.
A number of equalization techniques to mitigate ISI exist, including: (a) maximum likelihood sequence estimation (MLSE), in which a dynamic programming algorithm is used to determine the most likely transmitted sequence given observations of the received noisy and ISI-corrupted sequence and knowledge of the channel impulse response coefficients; (b) sub-optimal equalizer structures such as a linear equalizer (LE), wherein one simple finite impulse response (FIR) filter is used to mitigate ISI, or a non-linear decision feedback equalizer (DFE) that, in addition to the feed-forward FIR filter, employs a feedback filter (FBF) on the previously detected symbols; and (c) multi-carrier modulation (MCM), wherein the spectrum of the frequency-selective channel is divided into a large number of parallel, independent and approximately flat sub-channels using an orthogonal transformation.
MLSE uses a sequence of received signal samples over successive symbol intervals to make decisions about the transmitted symbols, and can be considered to be optimal from a bit error rate (BER) perspective. However, MLSE has a computational complexity that grows exponentially with the length of the channel time dispersion, and is typically prohibitively expensive to implement. In sub-optimal structures such as LE and DFE, data detection is done on a symbol-by-symbol basis and hence is much simpler to implement than the optimal MLSE. Linear equalization uses a linear filter with adjustable coefficients. Decision feedback equalization exploits a FBF to suppress that part of the ISI from the present estimate that was caused by previously detected symbols.
Advanced equalization techniques are usually too complex to be implemented using analog circuit design. Furthermore, even in the case of plain baseband communications that employ a simple LE, digital implementations are advantageous because of easy scalability to integrated circuit (IC) technologies of smaller dimensions. For this reason, most of the modern data receiver designs are digital, and a high-level description is illustrated in FIG. 1.
The received signal passes through the analog-front-end (AFE) block 102 that may include functions such as analog filtering, variable gain amplification (VGA), analog demodulation etc., depending on the application. Then, the signal is digitized in an analog-to-digital (ADC) conversion block 104. Finally, the digital signal samples are processed in a DSP receiver 106 that may include timing recovery, equalization and other types of advanced signal processing techniques.
Non-linearity of the overall communications channel from the data source to the DSP receiver 106 is one of the problems that can severely distort the signal and hence degrade the overall system performance. Examples of severe non-linearity are harsh signal compression and clipping.
A digital equalizer is typically very sensitive to a harsh non-linear distortion of the received signal. As an illustration, clipping can significantly reduce an equalizer's performance unless clipping occurs relatively infrequently, such as in less than 1% of the data transmission.
When dealing with the problem of equalizing clipped signals in MCM systems, some techniques are used to remove the effect of clipping by restoring the signal linearity through interpolation (see U.S. Pat. No. 6,606,047 to Borjesson, et al.). Or, some other algorithms are proposed to reduce the signal peak-to-average power ratio (PAPR) (see U.S. Pat. No. 7,340,006 to Yun, et al.).
In a wire-line application, one technique uses an equalizer on the transmit side and clips the transmit signal when overshoot occurs due to filtering (see U.S. Pat. No. 6,452,975 to Hannah).
Many digital communication systems attempt to avoid signal clipping. Examples of those approaches can be found in U.S. Pat. No. 7,336,729 to Agazzi and U.S. Pat. No. 7,346,119 to Gorecki, et al., both of which are applicable to high-speed serializer-deserializer (SerDes) digital designs.
One application of SerDes is to transfer data over backplane channels in chip-to-chip communications. An example of a channel impulse response (CIR) of a typical backplane at 6 Gbit/s is illustrated in FIG. 2.
The cursor level of the illustrated CIR is at 486.1 millivolts (mV), while its dominant post-cursor component is at 183.7 mV.
The following illustrative example is based on non-return to zero (NRZ) signaling that is relatively popular in SerDes applications. Under the assumptions of no correlations among the transmitted data, a Gaussian noise environment, and no non-linear distortions, the illustrated channel can be readily equalized with a linear minimum mean-squared-error (MMSE) equalizer because the MMSE equalizer is designed to minimize the slicer error variance and consequently bit-error-rate (BER).
The channel described earlier in connection with FIG. 2 does not introduce a severe ISI distortion, and hence, a relatively simple 2-tap equalizer having values as computed from CIR using the formulas can be used:
                              c          0                =                                                            h                0                            ·                                                ∑                  n                                ⁢                                  h                  n                  2                                                      -                                          h                                  -                  1                                            ·                                                ∑                  n                                ⁢                                  (                                                            h                                              n                        -                        1                                                              ·                                          h                      n                                                        )                                                                                                        (                                                      ∑                    n                                    ⁢                                      h                    n                    2                                                  )                            2                        -                                          (                                                      ∑                    n                                    ⁢                                      (                                                                  h                                                  n                          -                          1                                                                    ·                                              h                        n                                                              )                                                  )                            2                                                          Eq        .                                  ⁢        1            
                              c          1                =                                                            h                                  -                  1                                            ·                                                ∑                  n                                ⁢                                  h                  n                  2                                                      -                                          h                0                            ·                                                ∑                  n                                ⁢                                  (                                                            h                                              n                        -                        1                                                              ·                                          h                      n                                                        )                                                                                                        (                                                      ∑                    n                                    ⁢                                      h                    n                    2                                                  )                            2                        -                                          (                                                      ∑                    n                                    ⁢                                      (                                                                  h                                                  n                          -                          1                                                                    ·                                              h                        n                                                              )                                                  )                            2                                                          Eq        .                                  ⁢        2            
In Equations 1 and 2 above, hn represent samples of CIR at baud rate, with h0 being the cursor and h-1 the precursor.
The optimal tap values of the example channel are [c0, c1]=[2.12, −0.80].
In high-speed SerDes designs, the equalizer taps are usually normalized so that the first tap is equal to 1. Then, a 2-tap post-cursor equalizer is usually called a single tap equalizer.
This normalization does not change the equalizer shape or overall system performance. In the following example, a normalized equalizer, but this does not limit the equalizer does not need to be normalized.
As a result, the normalized tap C1 can be found from Equations 1 and 2 as:
                              c          1          ′                =                                                            h                                  -                  1                                            ·                                                ∑                  n                                ⁢                                  h                  n                  2                                                      -                                          h                0                            ·                                                ∑                  n                                ⁢                                  (                                                            h                                              n                        -                        1                                                              ·                                          h                      n                                                        )                                                                                                        h                0                            ·                                                ∑                  n                                ⁢                                  h                  n                  2                                                      -                                          h                                  -                  1                                            ·                                                ∑                  n                                ⁢                                  (                                                            h                                              n                        -                        1                                                              ·                                          h                      n                                                        )                                                                                        Eq        .                                  ⁢        3            
The optimal single tap post-cursor equalizer of the foregoing example is [1, −0.38].
If the received and equalized signal samples are denoted as xn and yn respectively, their relation in the normalized equalizer form is given in Equation 4:
                              y          n                =                              x            n                    +                                    c              1              ′                        ·                          x                              n                -                1                                                                        Eq        .                                  ⁢        4            
FIG. 3 illustrates received and equalized signal traces. FIG. 3 illustrates that equalizing clipped signals can be ineffective.